Finding a line with the following criteria

Question

Find the line that goes through $\ (0,1,2) $, is parallel to the plane $\ x+y+z-2=0 $, and is perpendicular to $\ r(t) = (1+t,1-t,2t) $.

I understand that the line is perpendicular to the vectors $\ (1,-1,2)$ and $(1,1,1) $, but I'd like some guidance as to what I should do.

I could only think of one cross product which is $\ (1,-1,2) \times (1,1,1) = 0 $ am I correct? — bm1125, Apr 07 '19 at 22:26
But isn't the cross product of two parallel vectors is zero? — bm1125, Apr 07 '19 at 22:35
The cross product of two parallel vectors is zero, but $(1,-1,2)$ and $(1,1,1)$ are not parallel (i.e., they are not scalar multiples of one another) — J. W. Tanner, Apr 07 '19 at 22:42
Oh okay. But if the vector that I look for is perpendicular to $\ (1,1,1) $ and also $\ (1,-1,2) $ doesn't that mean that those two vectors are parallel? — bm1125, Apr 07 '19 at 22:46
Not in $3$ dimensions; for example $\hat k$ is perpendicular to $\hat i $ and to $\hat j$, but $\hat i$ and $\hat j$ are not parallel — J. W. Tanner, Apr 07 '19 at 22:49
Thanks a lot for your explanations!! I'm having a hard time grasping it. — bm1125, Apr 07 '19 at 22:52

score 1 · Accepted Answer · answered Apr 07 '19 at 22:39

To find a vector in $\mathbb R^3$ perpendicular to two other linearly independent vectors, take the cross product, which can be computed as a formal determinant:

$$\left|\begin{matrix} \hat i &\hat j & \hat k\\ 1 & -1 & 2 \\ 1 & 1 & 1 \end{matrix}\right| = -3\hat i + \hat j + 2\hat k .$$

If you like, you could verify using the dot product that $(-3,1,2)$ is perpendicular to the other two vectors.

So the line you're seeking is given by $l(t)=(0,1,2)+t(-3,1,2)=(-3t,1+t,2+2t)$.

Finding a line with the following criteria

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